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New PDF release: Analytic Trigonometry

By William J. Bruce, W. J. Langford, E. A. Maxwell and I. N. Sneddon (Auth.)

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Prove: csc Θ = COS 0 cot 0 + tan 0 Since the right member is a cosine, we choose to express all functions in the left member in terms of sines and cosines and 30 ANALYTIC TRIGONOMETRY then apply the rules of elementary algebra to simplify the complex fraction obtained. 2 Express in a simplified form: 1. sin Θ csc 0. 2. cot Θ sin 0. 3. csc 0 sec Θ' 4. (sec 0 - 1) (sec 0 + 1). 5. (1 + sin Θ) (1 - sin Θ). 6. sec2 Θ sec2 0 - 1' 7. (1 + cot2 0) cos2 0. 8. cos 0 sin 0 sin 0 cos 0 * 9. (sin 0 + cos 0)2.

Express cos 4 Θ in terms of sin 2 0. cos 4 0 = 1 - 2 sin2 2 0. Example 2. Express sin2 2 0 in terms of cos 4 0. sin2 2 (9 = - (1 - cos 4 0). 56 ANALYTIC TRIGONOMETRY Example 3. Express cos4 Θ in first powers. cos4 0 = cos2 θ cos2 Θ = - (1 + cos 2 Θ) - (1 + cos 2 Θ) = - (1 + 2 cos 2 0 + cos2 2 0) = ~ | l + 2 cos 2 0 + ί (1 + cos 4 0)1 - I f 3 + 2 cos 2 0 + 1- cos 4 N0 . 2 1. Express in functions of half the angle: (a) sin 4 Θ. (b) sin k Θ (c) cos 6 0, in sines. (d) cos 3 0, in cosines. (e) sin 3 Θ.

Since sin2 0=1— cos2 0, this equation may be written 5 cos2 0 + 2 (1 - cos2 0) = - . Simplifying, we get 3 cos2 0 = — Thus, cos 0 = ± - V3~ 2 Since cos 0 is both positive and negative, 0 can be in any quadrant. The reference angle is 30° ( — ). Therefore, 0 = . . , - 30°, 30°, 150°, 210°, 330° Just as in solving algebraic equations, a word of caution is necessary in the solutions of trigonometric equations. Division by any factor that could be zero in an equation will result in the loss of certain solutions.

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